"Calculate the cell potential Ecell at 25oC for the cell if the initial concentration of Ni(NO3)2 is 0.100 molar and initial concentration of AgNO3 is 1.00 molar. [ EoNi2+/ Ni = -0.25V; EoAg+/Ag= 0.80V]"

1.0795V

E_cell = E^o_cell - \(\frac{0.059}{2}\) log\(\frac{[Ni^{2+}]}{[Ag^+]^2}\) 

E^o_cell = 0.80 + 0.25 = 1.05 V
E_cell = 1.05 - \(\frac{0.059}{2}\) log\(\frac{0.1}{1^2}\)

=1.05 - \(\frac{0.059}{2}\) x (-1) =1.05 + 0.0295

E_cell = 1.0795 V